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Longer Vector Algebra Proof

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data
    this is an Apostol problem in chapter 12 and I guess it's a hypothetical definition of a norm of a vector
    Assuming this different definition of the norm prove these statements-
    [tex]Def. ||A||=\sum_{k=1}^{n}|a_{k}|, prove ||A||>0, if ||A||\neq0,||A||=0 if A=0, ||cA||=c||A||,triangle equality ||A+B||\leq ||A||+||B||,[/tex]

    I just looked at this- do I just expand the sums out and then express them in sigma notation?

    Use this definition in V2 and prove on a figure the set of all points (x,y) of norm 1- this is just the line x+y=1?

    which of the above theorems/statements would hold if we take the absolute value of the summation?

    [tex] Def. ||A||=|\sum_{k=1}^{n}a_{k}| [/tex]


    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 24, 2013 #2

    LCKurtz

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    You haven't told us what ##A## is or what the ##a_k## are. If ##n = 2## is ##v = \langle 1,-1\rangle \in A##? If so ##v \ne 0## but ##\|v\| = 0##.
     
  4. Oct 24, 2013 #3
    Mostly I'm asking about the triangle inequality proof, the problem just gives you a general vector A
    I think the hypothetical definition is ||A||=|a1|+|a2|+|a3|...+|an|, so if ||A|| norm doesn't equal the zero vector (0,0...0) and at least one component |ak|>0, then the sum must be >0
     
  5. Oct 24, 2013 #4

    LCKurtz

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    You "think" that is the norm definition? It's your question -- is it or isn't it? I see you edited your post to change it to that. And given your last sentence, what is your question?

    [Edit] Re the triangle inequality, what have you tried?
     
  6. Oct 25, 2013 #5
    I'm just trying the TE for the first time. Apostol gives this 'incorrect' definition of the norm and asks you to prove the TE. I proved the law of cosines in norm form. here I see it is correct for a case where the signs of A and B are different

    [tex]Def. ||A||=\sum_{k=1}^{n}|a_{k}|, triangle.inequality ||A+B||\leq ||A||+||B||. e.g.A=-2,B=1,|-1|\leq 3[/tex]
     
  7. Oct 25, 2013 #6

    LCKurtz

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    What do you mean the signs of A and B? That doesn't make any sense. They are vectors, not scalars. You have to start by calculating A+B and its norm and compare it with the norms of A and B.
     
  8. Oct 25, 2013 #7
    I've attached the written problems, is this how I prove the triangle equality for this definition of norm, I ended up with something that seems analogous to Cauchy-Schwarz inequality?

    anyways I started to lose confidence after thinking my handwriting was too messy after I saw a video of Andrew Wiles
     

    Attached Files:

    Last edited: Oct 25, 2013
  9. Oct 25, 2013 #8

    LCKurtz

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    You are making it way too hard. Like I said before, calculate A+B and its norm. You don't need to square anything. Just use a familiar inequality. Show your work here, not in a pdf.
     
  10. Oct 25, 2013 #9
    [tex]||A+B||\leqslant ||A||+||B||\rightarrow \sum_{k=1}^{n}|ak+bk|\leq \sum_{k=1}^{n}|ak|+\sum_{k=1}^n|bk|[/tex]
    so this is trivially true by properties of absolute value? there are many more beautiful properties of absolute value functions

    the last part of the solution is
    |x|+|y|=1 and y=1-x, y=x+1, y=x-1, y=-x-1, and since |x|+|y|=1, then these lines become segments, 0<=x<=1 constrained and form a rectangle in V2 (similar to R2?)
     
    Last edited: Oct 25, 2013
  11. Oct 25, 2013 #10

    LCKurtz

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    No, that is so wrong it leaves me speechless. One more time I will ask you:

    1. Write down A = ... and B = ... and then what is A + B is.
    2. Then write down its norm ##\|A+B\| =##?
    Then you are ready to try to get the inequality.
     
  12. Oct 25, 2013 #11
    Sorry I'm not sure what you mean, the text gives you a hypothetical definition of norm as the sum of absolute values of |ak|
    the true definition of norm as dot product gives a proof of the triangle equality from algebra or as a consequence of the Cauchy-Schwarz inequality
     

    Attached Files:

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  13. Oct 25, 2013 #12

    LCKurtz

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    Sorry, but I will not download a pdf file for a simple problem like this. You can easily type it. But until you do 1. and 2. above, I don't see where we have anything to discuss.

    The Cauchy-Schwarz inequality has nothing to do with this problem.

    [Edit] I now see you have changed post #9 to make it correct, without noting that it was changed. That makes it very difficult to maintain continuity of the thread and is against forum rules.
     
    Last edited: Oct 25, 2013
  14. Oct 28, 2013 #13
    [tex]A=\sum_{k=1}^{n}akEk, B=\sum_{k=1}^{n}bkEk, ||A||=\sqrt{A\cdot A}.
    \\ ||A+B||=\sum_{k=1}^{n}[(ak+bk)^{2}]^{1/2},
    \\ but here ||A||=\sum_{k=1}^{n}|ak|, ||B||=\sum_{k=1}^{n}|bk|,
    \\ ||A||+||B||=\sum_{k=1}^{n}(|ak|+|bk|), ||A+B||=\sum_{k=1}^{n}\|ak+bk|
    \\ \sum_{k=1}^{n}\|ak+bk| <=\sum_{k=1}^{n}(|ak|+|bk|) [/tex]


    I'm sorry, so all vectors in Vn can be expressed as linear combinations of the orthogonal basis vectors [Ek,En] or other arbitrarily rotated orthogonal basis vectors as well as the imaginary basis vector on the C plane? I've written what I think is the definition of norm/length and also the hypothetical one (I'm not sure if there's some significance to Apostol's hypothetical problem)
     
    Last edited: Oct 28, 2013
  15. Oct 28, 2013 #14
    actually since ||A|| is defined as the sum of absolute values then if v=(1,-1) then ||v||=2, not 0, while normally
    [tex]||v||=2^{1/2}[/tex]
    it is really the greatest achievement of humanity! by the way what would be an example of a difficult proof to try that is more advanced or a text?
     
    Last edited: Oct 28, 2013
  16. Nov 6, 2013 #15
    am I correct in the last post?
     
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